A Universally derided "primality test".

[RMLabrador]

Ok. Once again, very well. Good, good explanation, that explain all and not explain nothing at the same time)) I'm try to explain, feel free to correct me - I'm not mathematician. Only month ago or so I'm read about modular arithmetic))
Just look at this. u=0 in my notation == The test is Fibonacci PRP test. If we write power of matrix as a*c-((c-a)/(u-1))^2==(u-1)^p, for u=0
a*c - ((c-a)/(-1))^2==(-1)^p. Let p be uneven, so a*c-(a-c)^2==-1
for u=2 (Lucas PRP))) -> a*c-(c-a)^2==1
Ok? Still no wise math word?)) Remember, that a,c - always integer!!!
Plot the implicit curves, WolframAlfa
https://www.wolframalpha.com/input/?...+0%2C+10%7D%5D
an what we see?
Two hyperbolae, and for any a we can easy compute the difference between c values
((5a^2+4)^.5-(5a^2-4)^.5)/2
Easy to see that this value never be an integer for any integer a.
I.e. for any a value form Fibonacci test, there not exist an integer value of c for Lucas test for all values of the p - p is not present in the right side of equation.
So Fermat PRP never match with Lucas PRP and wise versa.
Its elementary and were here is my mistake?

[Dr Sardonicus]

I don't understand what you're asking. You are looking at det(A^p), which is (det(A))^p = (u-1)^p.

We have A = [1,1;1,u], Aⁿ = [a_n(u), b_n(u); b_n(u), c_n(u)] for integer n. The polynomials a, b, c have integer coefficients (easily proved by induction).

The formula (c_n(u) - a_n(u))/(u-1) = b_n(u) is correct (easily proved by induction). Congratulations. The polynomial identity

a_n(u)*c_n(u) - b_n²(u) = (u-1)ⁿ

is then easily shown to be equivalent to the identity I gave earlier, [recall a_n(u) + c_n(u) = L_n, b_n(u) = F_n, Δ = (u-1)^2 + 4]

L_n² - Δ*F_n² = 4*(u-1)ⁿ.

We have, for p prime, u in Z/pZ, A^p == A (mod p) if (Δ/p) = +1; and

A^p == [u+1,0;0,u+1] - A = [u,-1;-1,1] (mod p) if (Δ/p) = -1.

In either case, the determinant of A^p is congruent to u-1 (mod p) for u in Z/pZ. For u = 0, 2 we have Δ = 5. Other than the case p = 5, we get

det(A^p) == 1 (mod p) for u = 2, and det(A^p) == -1 (mod p) for u = 0. I don't see what the problem is.

[RMLabrador]

)) There is no problem! Please, read my post above, this IS the proof about Fermat and Lucas probable prime do not inteecept or not? Its important, as far as I'm too do not understead, that I'm write understadeble)))
You stated that Fermat and Lucas test are PRP, so all their combination are PRP too. Thats right if use modulo form only, and have no care about existance of factor in this polynomial. Factor inevery their coefficient lead as to symmetry, the ones is key to proof that even in modulo form the correct, non-prp test can be built.
I need somehow post on arxive.org, can i do this without invitation?

[Dr Sardonicus]

Originally Posted by RMLabrador

There is no problem! Please, read my post above, this IS the proof about Fermat and Lucas probable prime do not inteecept or not?
<snip>

I don't know what the heck you're trying to say.

If you mean that (Δ/p) = -1 and (Δ/p) = +1 never occur simultaneously for a given Δ and p > 2, that is trivial.

[RMLabrador]

Originally Posted by RMLabrador

I.e. for any a value form Fibonacci test, there not exist an integer value of c for Lucas test for all values of the p - p is not present in the right side of equation.
So Fermat PRP never match with Lucas PRP and wise versa.

Certainly, I mean, Fermat and Lucas Pseudoprimes do not intercept at any p

[Dr Sardonicus]

Originally Posted by RMLabrador

Certainly, I mean, Fermat and Lucas Pseudoprimes do not intercept at any p

I don't think you know what you're talking about.

[kruoli]

Originally Posted by Dana Jacobsen, A005845

For example, the number 82380774001 is both an A005845 Lucas pseudoprime and a Fermat pseudoprime to the first 407 prime bases. - Dana Jacobsen, Jan 10 2015

Yes...

[RMLabrador]

Fibonacci vs Lucas))) Stupid me - 100%. I'n even correct the post, in the post, were link with hyperbolae, started as Fibonacci. Please, exuse me.

[CRGreathouse]

Grantham has an explicit finite set he conjectures to contain a number which is both a Carmichael number and a Lucas pseudoprime, see A018188.

[Uncwilly]

Originally Posted by RMLabrador

Someone even quietly, without notice, changed the name of the topic)))))

Originally Posted by LaurV

Prove him an a******, by coming with a proof of the fact that your test is more than a PRP test. Otherwise you are just a crank or (worse) a troll.

Originally Posted by Dr Sardonicus

As to changing thread titles, it's a common occurrence on this forum. It's childish, but acting childishly does not make someone an a.

You may appeal via PM to any Super-mod (those whose names are in red) you choose to, to have them change the title again (you may not like the result of that). Each has the power to change it. Moving the thread to Misc Math was one of the moderators ways of expressing their opinion about your primality test. Changing the title is another way a moderator can express their opinion, so as to warn others up front.

[CRGreathouse]

Originally Posted by Uncwilly

You may appeal via PM to any Super-mod (those whose names are in red) you choose to, to have them change the title again (you may not like the result of that). Each has the power to change it. Moving the thread to Misc Math was one of the moderators ways of expressing their opinion about your primality test. Changing the title is another way a moderator can express their opinion, so as to warn others up front.

I move many threads to Misc Math. I haven't yet abused my power to rename threads (though I have occasionally renamed threads with vague or misleading titles to be more explanatory, albeit bland).